Timeline for Growth of the "denominator" of powers of an algebraic number
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2013 at 3:02 | comment | added | Will Sawin | @QiaochuYuan: Take the ideal $I$ of algebraic integers $a\ in \mathbb Q(z)$ such that $az$ is an integer. $I$ is a proper nontrivial ideal, so it has some proper nontrivial prime factor. This defines a valuation. The valuation of $z$ by that valuation cannot be nonnegative, or else $z$ could be written as $b/a$ for $b$ an algebraic integer and $a$ a unit mod the prime ideal, which is a contradiction because $a$ is not in $I$. | |
| Nov 22, 2013 at 1:55 | vote | accept | Terry Tao | ||
| Nov 22, 2013 at 1:20 | comment | added | Qiaochu Yuan | The point should be that integrality is a local condition ($z$ is an algebraic integer iff $\nu(z) \ge 0$ for every valuation $\nu$ on $\mathbb{Q}(z)$) and I feel like there should be a clean proof of this but I don't know it. | |
| Nov 22, 2013 at 1:10 | answer | added | Julian Rosen | timeline score: 24 | |
| Nov 22, 2013 at 0:54 | history | asked | Terry Tao | CC BY-SA 3.0 |